How do I solve this dynamics problem involving a pulley?

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Homework Help Overview

The discussion revolves around a dynamics problem involving a pulley system, where participants are analyzing forces acting on blocks, including tension and friction. The original poster expresses difficulty in setting up the free body diagrams (FBDs) and determining the correct signs for acceleration and tension.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of free body diagrams and the assignment of positive and negative directions for forces. There are attempts to clarify the relationship between tension and the direction of the applied force. Some participants question the consistency of tension values across different FBDs and the implications of choosing different coordinate systems.

Discussion Status

There is ongoing exploration of the equations governing the system, with some participants providing guidance on how to approach the calculations. Multiple interpretations of the signs for tension and acceleration are being examined, and participants are working through their calculations to reconcile discrepancies in their results.

Contextual Notes

Participants mention specific values for friction and normal forces, indicating that the problem involves multiple surfaces and forces acting on the blocks. There is a recognition of the need to account for the total mass when calculating normal forces, which adds complexity to the problem.

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Homework Statement



http://14.photobucket.com/albums/a322/guitaristx/phy.jpg

Just add the http for the image above..

Homework Equations



Please let me know how the FBD looks and I think I am having trouble with my coordinate system( direction and signs)

The answer for acceleration is 5.21m/s^2...and the tension is 215N...I can't seem to get the equation to work out to the correct answer...any help and direction would be appreciated...

The Attempt at a Solution



http://i14.photobucket.com/albums/a322/guitaristx/Scan10005.jpg

just add the http for the image above...



 
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Jason03 said:
Please let me know how the FBD looks and I think I am having trouble with my coordinate system( direction and signs)
Your FBD looks OK. Pick a direction parallel to the plane as positive and stick to it, such as: Down = positive; up = negative.
 
I chose up as positive and down as negative and for the equations that I have for the first FBD...and the only way I come up with the correct Tension answer is by making the acceleration negative...and for the second FBD my calculations came up with a completely different tension...when the tensions should obviously be the same
 
Jason03 said:
I chose up as positive and down as negative and for the equations that I have for the first FBD...
Then why is tension marked -T?
 
The tension is upward...but the problem states that P is a force pulling block A downward...so I figured Tension would be negative because block A is being pulled downward(left) hence the Negative...
 
Jason03 said:
The tension is upward...but the problem states that P is a force pulling block A downward...so I figured Tension would be negative because block A is being pulled downward(left) hence the Negative...
I don't quite understand that reasoning. The tension is upward and thus positive. (Ropes can't push--only pull.)
 
Jason03 said:
Im not sure how to explain it...but here's an example of an upward Tension on a block going down that has a negative T
Only because they chose down to be positive.
 
ok...I think i see now what you mean...let me try some new calculations...
 
  • #10
i changed the T to positive...but when I plug in 5.21 for the acceleration I get a -215 for the Tension??
 
  • #11
Show me the equations you used.

Don't forget that the acceleration will be negative (down the incline).
 
Last edited:
  • #12
heres the equ.

T-P+Fr-mgSin(30) = ma

I tried the neg. accel. and it gave me an answer even further off...
 
  • #13
As long as you have the acceleration negative, that equation should work out OK. What numbers did you plug in? What did you use for Fr?
 
  • #14
Thanks for the help on the first one, it finally worked out...

But for the second FBD I came up with

T-Fr-mgSin30 = ma

But for Friction there are two surfaces. So I have to find both normal forces acting on block A and block B

heres the numbers I came up with

Friction for B = .10(15)(9.8)Cos(30)= 12.7N

Friction for A = .10(30)(9.8)Cos(30)= 25.4N

So

T-12.7N-25.4-73.5N = 15 (5.21)

the acceleration wouldn't be negative now would it?

but now T is not coming out to 215 as it should...do you see any errors in my calculations?
 
  • #15
Jason03 said:
But for the second FBD I came up with

T-Fr-mgSin30 = ma
Looks good.
But for Friction there are two surfaces. So I have to find both normal forces acting on block A and block B

heres the numbers I came up with

Friction for B = .10(15)(9.8)Cos(30)= 12.7N

Friction for A = .10(30)(9.8)Cos(30)= 25.4N
Careful: What's the normal force of the bottom block against the incline?
 
  • #16
ohh.. I did miss that, the total mass of both blocks acting as N on the lower block.

The numbers worked out.

So to solve the system I am trying this

T-372 = 30(-a)
T-177 = 15 (a)


So can I cancel the acceleration to find T?..or is there a better way...
 
  • #17
*T-137 = 15a
 
  • #18
Got it!...thanks for all the help:approve:
 

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